How do you find the volume of the pyramid bounded by the plane 2x+3y+z=6 and the coordinate plane?

Question

Christopher Agresta · Accepted Answer

#= 6 # cubic units

the normal vector is #((2),(3),(1))# which points out in the direction of octant 1, so the volume in question is under the plane and in octant 1

we can re-write the plane as #z(x,y)= 6 - 2x - 3y #

for #z = 0# we have

#z= 0, x = 0 implies y = 2#
#z= 0, y = 0 implies x = 3#

and
- - #x= 0, y = 0 implies z = 6#

it's this:

the volume we need is

#int_A z(x,y) dA#

#= int_(x=0)^(3) int_(y=0)^(2 - 2/3 x) 6 - 2x - 3y \ dy \ dx#

#= int_(x=0)^(3) [ 6y - 2xy - 3/2y^2 ]_(y=0)^(2 - 2/3 x) \ dx#

#= int_(x=0)^(3) [ 6(2-2/3 x) - 2x(2-2/3 x) - 3/2(2-2/3 x)^2 ]_(y=0)^(2 - 2/3 x) \ dx#

#= int_(x=0)^(3) 12-4 x - 4x + 4/3 x^2 - 6 - 2/3 x^2 + 4x \ dx#

#= int_(x=0)^(3) 6- 4 x + 2/3 x^2 \ dx#

#=[ 6x- 2 x^2 + 2/9 x^3 ]_(x=0)^(3)#

#= 18- 18 + 54/9 #

#= 6 #

Evelyn Agard · Answer

6

We are going to be performing a triple integral. The cartesian coordinate system is the most applicable. The order of integration is not critical. We are going to go z first, y middle, x last. #underline("Determining limits")# On the plane #z = 6 - 2x - 3y# and on the coordinate plane #z = 0# hence # z: 0 rarr 6 - 2x - 3y# Along #z=0#, #y# goes from 0 to #3y = 6 - 2x# hence #y: 0 rarr 2 - 2/3x# Along #y=0, z=0# hence #x: 0 rarr 3# We are finding the volume so #f(x,y,z) = 1#. Integral becomes #int_0^3int_0^(2-2/3x)int_0^(6-2x-3y)dzdydx# #=int_0^3int_0^(2-2/3x)[z]_0^(6-2x-3y)dydx# #=int_0^3int_0^(2-2/3x)(6-2x-3y)dydx# #=int_0^3[6y-2xy - 3/2y^2]_0^(2-2/3x)dx# #=int_0^3(6(2-2/3x) - 2x(2-2/3x) - 3/2(2-2/3x)^2) dx# #=int_0^3 (12 - 4x - 4x + 4/3x^2 - 3/2(4 - 8/3x + 4/9x^2)) dx# #=int_0^3 (12 - 8x + 4/3x^3 - 6 + 4x - 2/3x^2)dx# #=int_0^3(6 - 4x + 2/3x^2)dx# # = [6x - 2x^2 + 2/9x^3]_0^3# #=6(3) - 2(3)^2 +2/9(3)^3 # #=6#

Samantha Adkison · Answer

undefined To find the volume of the pyramid bounded by the plane $2x + 3y + z = 6$ and the coordinate plane, you first need to determine the height of the pyramid. The height is the perpendicular distance from the apex of the pyramid (the point where the three edges meet) to the base.

The equation of the plane can be rewritten in the form $z = 6 - 2x - 3y$. Since the plane intersects the coordinate axes, you can find the intercepts by setting $x = 0$, $y = 0$, and $z = 0$ respectively.

Setting $x = 0$, you find $z = 6$, which is the intercept on the z-axis. Setting $y = 0$, you find $z = 6 - 2x$, which gives the intercept on the x-axis. Setting $z = 0$, you find $y = 2$, which gives the intercept on the y-axis.

From the intercepts, you can determine that the base of the pyramid is a triangle with vertices at (0, 0, 6), (0, 2, 0), and ($3, 0, 0$).

To find the height of the pyramid, you need to find the distance from the apex to the base. This can be calculated using the distance formula or by finding the distance between the apex and a point on the base.

Using the distance formula, the height $h$ is given by $h = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$, where $(x_1, y_1, z_1)$ is the apex and $(x_2, y_2, z_2)$ is a point on the base.

Substituting the coordinates, you can calculate the height.

Once you have the height and the area of the base triangle, you can use the formula for the volume of a pyramid:

$$ V = \frac{1}{3} \times \text{Base Area} \times \text{Height} $$

Substitute the values to find the volume of the pyramid.